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UNIT 1A LESSON 2

UNIT 1A LESSON 2. FACTORING POLYNOMIAL EXPRESSIONS. Factor Review. x 2 + 3 x – 4 . x 2 – 25 . ( x – 5)( x + 5). ( x + 4)( x – 1). x 2 – 9 x + 8 . 4 x 2 – 20 x + 25. ( x – 8)( x – 1). (2 x – 5) 2. 2 x 2 – x – 3 . 3 x 2 + x – 4. 3 x 2 – 3 x + 4 x – 4

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UNIT 1A LESSON 2

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  1. UNIT 1ALESSON 2 FACTORING POLYNOMIAL EXPRESSIONS

  2. Factor Review x2 + 3x – 4 x2 – 25 (x – 5)(x + 5) (x + 4)(x – 1) x2 – 9x + 8 4x2 – 20x + 25 (x – 8)(x – 1) (2x – 5)2 2x2 – x – 3 3x2 + x – 4 3x2 – 3x + 4x – 4 3x(x – 1) + 4(x – 1) (3x + 4)(x – 1) 2x2 + 2x – 3x – 3 2x(x + 1) – 3(x + 1) (2x – 3)(x + 1)

  3. Factoring Higher Degree Polynomials Do you remember how to factor polynomials such as F(x) = x3 + 6x – 4x + 1 G(x) = x4 – 7x2 + 5x – 2 or if they even factor?

  4. Factor by Grouping Sometimes, a polynomial expression can be factored by grouping. (Similar to decomposition) • Example 1 2x3+ 3x2 – 8x – 12 x2(2x + 3) – 4 (2x + 3) (x2 – 4)(2x + 3) (x – 2)(x + 2)(2x + 3)

  5. Practise 1. f(x) = x3 + x2– 9x – 9 x3 + x2 – 9x – 9 x2(x + 1) – 9 (x + 1) (x2 – 9)(x + 1) (x – 3)(x + 3)(x + 1) 2. g(x) = 4x3 + 8x2 – 25 x – 50 4x3 + 8x2 – 25x – 50 4x2(x + 2) – 25 (x + 2) (4x2 – 25)(x + 2) (2x – 5)(2x + 5)(x + 2)

  6. Complete Question 1

  7. Factoring Polynomial Expressions EXAMPLE 2: Consider the polynomial P(x) = x3 – 7x– 6 Step 1: List the integers, which divide into the constant term–6 These are called potential zeros + 1, +2, +3, +6 If P(a) = 0 then (x – a) is a factor Step 2: Test the potential zeros until you find zero then state the factor P(1) = (1)3 – 7 (1) – 6 = –12 therefore ( x – 1) is NOT a factor P(–1) = (–1)3– 7 (–1) – 6 = 0 therefore ( x+ 1) is IS a factor

  8. Coefficients of P(x) • 1 0 -7 -6 • 1 -1 -6 • 1 -1 -6 0 Synthetic Division P(x) = x3 – 7x – 6 has the factor x + 1 Step 3: Divide the polynomial with the factor you found Method 1: Using subtraction Quotient x2 – x – 6 Remainder 0 Step 4: Complete the factoring of the polynomial P(x) = x3 – 7x – 6 P(x) = (x + 1)(x2 – x – 6) P(x) = (x + 1)(x – 3)(x + 2)

  9. (-2, 0) (3, 0) (-1, 0) Check by Graphing Graph the polynomial P(x) = x3 – 7x – 6 The x-intercepts give us the zeros of the function which we can use to determine the factors (x + 2)(x + 1)(x – 3)

  10. Complete questions 2-3 on pages 3-4

  11. EXAMPLE 3: P(x) = x3 – x2 – 5x + 6 Find the zeros of P(x) by graphing and determining the x-intercepts 2 –2.3 1.3 Integral Zero ApproximateZeros

  12. -2 1 -1 -5 6 -2 -2 6 1 1 -3 0 EXAMPLE 3: P(x) =x3 – x2 – 5x + 6 From the previous graph the Integral zero is x = 2 OR P(2) = (2)3 – (2)2 – 5(2) + 6 =0so (x – 2) is a factor P(x) = (x – 2)(x2 + x – 3) Exact roots x = 2, x = – 2.3 or x = 1.3 Approximate roots

  13. Complete question 4 page 6

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